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Gaseous combustion products with the molar analysis of \(15 \% \mathrm{CO}_{2}, 25 \% \mathrm{H}_{2} \mathrm{O}, 60 \% \mathrm{~N}_{2}\) enter an engine's exhaust pipe at \(1100^{\circ} \mathrm{F}, 1 \mathrm{~atm}\) and are cooled as they pass through the pipe, to \(125^{\circ} \mathrm{F}, 1 \mathrm{~atm}\). Determine the heat transfer at steady state, in Btu per lb of entering mixture.

Short Answer

Expert verified
-9771.7 Btu/lb, indicating heat loss.

Step by step solution

01

- Convert temperatures to Rankine

First, convert the given temperatures from Fahrenheit to Rankine. Use the formula: \[T_R = T_F + 459.67\].
02

- Molar Heat Capacity Data

Use the molar heat capacity values at constant pressure for each component. Approximate values at standard conditions (1 atm) are: \[C_{p,\text{CO}_2} \approx 8.92 \frac{Btu}{lb \bullet°R} \] \[C_{p,\text{H}_2O} \approx 18.02 \frac{Btu}{lb \bullet°R} \] \[C_{p,\text{N}_2} \approx 6.96 \frac{Btu}{lb \bullet°R} \]
03

- Determine Average Molar Heat Capacity, \(\overline{C_p}\)

Calculate the average molar heat capacity for the gas mixture using the molar percentage of each gas: \[\overline{C_p} = y_{\text{CO}_2}C_{p,\text{CO}_2} + y_{\text{H}_2O}C_{p,\text{H}_2O} + y_{\text{N}_2}C_{p,\text{N}_2}\] Using the given molar composition (\(y_{\text{CO}_2}\) = 0.15, \(y_{\text{H}_2O}\) = 0.25, \(y_{\text{N}_2}\) = 0.60): \[\overline{C_p} = (0.15)(8.92) + (0.25)(18.02) + (0.60)(6.96)\]
04

- Calculate \(\overline{C_p}\) Value

Substitute the values to find \(\overline{C_p}\): \[\overline{C_p} = 1.338 + 4.505 + 4.176 = 10.019 \frac{Btu}{lb \bullet°R} \]
05

- Calculate Temperature Change in Rankine

Find the temperature change using the converted temperatures: \[\Delta T = T_{2R} - T_{1R} = 125 + 459.67 - (1100 + 459.67) = 125 - 1100 = -975 °R\] The temperature decreases by 975 °R.
06

- Calculate Heat Transfer per Pound Mixture

Use the formula for heat transfer: \[q = \overline{C_p} \Delta T\] Substitute the values: \[q = 10.019 \times (-975) = -9771.7 \frac{Btu}{lb} \] The negative sign indicates heat loss.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

steady state
The term 'steady state' refers to a condition where the properties of a system do not change with time. In thermodynamics and heat transfer, steady state implies that the temperature, pressure, and other thermodynamic properties remain constant over time. This condition simplifies the analysis because it allows us to focus only on the changes happening within the system without worrying about time-dependent variables. For this exercise, the exhaust pipe is assumed to be in a steady state, meaning the heat entering and leaving the pipe remains consistent.
molar heat capacity
Molar heat capacity, indicated as \( C_p \), is the amount of heat required to raise the temperature of one mole of a substance by one degree. It is important when analyzing the heat transfer in gas mixtures. Each component in the mixture has its own molar heat capacity value. For instance, \( C_{p, CO_2} = 8.92 \frac{Btu}{lb \bullet °R} \) for carbon dioxide, while water vapor, \( C_{p, H_2O} = 18.02 \frac{Btu}{lb \bullet °R} \), and nitrogen gas, \( C_{p, N_2} = 6.96 \frac{Btu}{lb \bullet °R} \). By calculating the molar heat capacity of the mixture using the provided molar fractions and the molar heat capacities of the individual gases, the overall heat transfer can be determined more accurately.
temperature conversion
Temperature conversion is essential in thermodynamics to maintain consistency across different units. In this exercise, we used Fahrenheit and Rankine. The temperatures given in Fahrenheit (°F) need to be converted to Rankine (°R) for the calculations. The conversion formula is simple: add 459.67 to the Fahrenheit temperature. For example, converting 1100°F to Rankine gives \[ T_{1100}^{\text{R}} = 1100 + 459.67 = 1559.67 \text{°R} \] and similarly, converting 125°F to Rankine gives \[ T_{125}^{\text{R}} = 125 + 459.67 = 584.67 \text{°R} \]. These conversions are crucial for further calculations like finding the temperature difference.
combustion products
Combustion products refer to the materials left after a combustion reaction occurs. In the given problem, the combustion products are a mixture with 15% carbon dioxide (CO2), 25% water vapor (H2O), and 60% nitrogen (N2). Analyzing these products helps in calculating heat transfer, as each gas has different properties that affect thermal calculations. Understanding the composition and behavior of these gases is necessary for accurate thermodynamic analysis, especially when dealing with engines and exhaust pipes, where efficient heat management is critical.

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Most popular questions from this chapter

An equimolar mixture of helium (He) and carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) enters an insulated nozzle at \(260^{\circ} \mathrm{F}, 5 \mathrm{~atm}\), \(100 \mathrm{ft} / \mathrm{s}\) and expands isentropically to a velocity of \(1110 \mathrm{ft} / \mathrm{s}\). Determine the temperature, in \({ }^{\circ} \mathrm{F}\), and the pressure, in atm, at the nozzle exit. Neglect potential energy effects.

At steady state, a device for heating and humidifying air has \(250 \mathrm{ft}^{3} / \mathrm{min}\) of air at \(40^{\circ} \mathrm{F}, 1 \mathrm{~atm}\), and \(80 \%\) relative humidity entering at one location, \(1000 \mathrm{ft}^{3} / \mathrm{min}\) of air at \(60^{\circ} \mathrm{F}\), 1 atm, and \(80 \%\) relative humidity entering at another location, and liquid water injected at \(55^{\circ} \mathrm{F}\). A single moist air stream exits at \(85^{\circ} \mathrm{F}, 1\) atm, and \(35 \%\) relative humidity. Using data from the psychrometric chart, Fig. A-9E, determine (a) the rate of heat transfer to the device, in Btu/min. (b) the rate at which liquid water is injected, in \(\mathrm{lb} / \mathrm{min}\). Neglect kinetic and potential energy effects.

A mixture having a molar analysis of \(60 \% \mathrm{~N}_{2}\) and \(40 \%\) \(\mathrm{CO}_{2}\) enters an insulated compressor operating at steady state at \(1 \mathrm{bar}, 30^{\circ} \mathrm{C}\) with a mass flow rate of \(1 \mathrm{~kg} / \mathrm{s}\) and is compressed to 3 bar, \(147^{\circ} \mathrm{C}\). Neglecting kinetic and potential energy effects, determine (a) the power required, in \(\mathrm{kW}\). (b) the isentropic compressor efficiency. (c) the rate of exergy destruction, in \(\mathrm{kW}\), for \(T_{0}=300 \mathrm{~K}\).

Outside air at \(50^{\circ} \mathrm{F}, 1\) atm, and \(40 \%\) relative humidity enters an air-conditioning device operating at steady state. Liquid water is injected at \(45^{\circ} \mathrm{F}\) and a moist air stream exits with a volumetric flow rate of \(1000 \mathrm{ft}^{3} / \mathrm{min}\) at \(90^{\circ} \mathrm{F}, 1 \mathrm{~atm}\) and a relative humidity of \(40 \%\). Neglecting kinetic and potential energy effects, determine (a) the rate water is injected, in \(\mathrm{lb} / \mathrm{min}\). (b) the rate of heat transfer to the moist air, in Btu/h.

A mixture of nitrogen and water vapor at \(200^{\circ} \mathrm{F}, 1 \mathrm{~atm}\) has the molar analysis \(80 \% \mathrm{~N}_{2}, 20 \%\) water vapor. If the mixture is cooled at constant pressure, determine the temperature, in \({ }^{\circ} \mathrm{F}\), at which water vapor begins to condense.

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